Chapter 11, Problem 62RE

a.

To determine

To find: A set of parametric equation of the lines passing through given point and parallel to the specified vector.

Answer to Problem 62RE

$x=3+t,y=2+t,z=1+t$

Explanation of Solution

Given information:

A line is passing through point $(3,2,1)$ and is parallel to line given by $x=y=z$

Concepts used:

Vectors are used for determining equation of a line in space. A line passing through the point $P(x_1,y_1,z_1)$ and parallel to the vector $\text{v}=\langle a,b,c\rangle$ is represented by the Parametric equations-

$x=x_1+at,y=y_1+bt,z=z_1+ct$ ;where, $a,b,c$ are direction numbers of vector $\text{v}=\langle a,b,c\rangle$ .

Calculation:

Line is passing through the point $(3,2,1)$ .

So, $x_1=3,y_1=2,z_1=1$ .

Line is parallel to the vector $x=y=z$ .

$\Rightarrow \frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{1}$

So, direction numbers are $a=1,b=1,c=1$ .

Now parametric equation of the lines passing through point $(3,2,1)$ and is parallel to $x=y=z$ will be given as-

$\begin{array}{l}x=x_1+at,y=y_1+bt,z=z_1+ct\\ \Rightarrow x=3+1t,y=2+1t,z=1+1t\\ \Rightarrow x=3+t,y=2+t,z=1+t\end{array}$

Hence, desired parametric equation of line is $x=3+t,y=2+t,z=1+t$ .

b.

To determine

To find: A set of symmetric equation of the line passing through given point and parallel to the specified vector.

Answer to Problem 62RE

$x-3=y-2=z-1$ .

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Concepts used:

Vectors are used for determining equation of a line in space. A line passing through the point $P(x_1,y_1,z_1)$ and parallel to the vector $\text{v}=\langle a,b,c\rangle$ is represented by the Parametric equations-

$x=x_1+at,y=y_1+bt,z=z_1+ct$ ;where, $a,b,c$ are direction numbers of vector $\text{v}=\langle a,b,c\rangle$ .

When direction numbers $a,b,c$ are non-zero then parameter $t$ can be eliminated to obtain set of symmetric equation $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

Calculation:

Line is passing through the point $(3,2,1)$ .

So, $x_1=3,y_1=2,z_1=1$ .

Since, direction numbers $a=1,b=1,c=1$ are all non-zero.

Therefore, set of symmetric equation will be given as −

$\begin{array}{l}\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \Rightarrow \frac{x-3}{1}=\frac{y-2}{1}=\frac{z-1}{1}\\ \Rightarrow x-3=y-2=z-1\end{array}$

Hence, desired set of symmetric equation of line is $x-3=y-2=z-1$ .